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A certain basket contains 10 apples, 7 of which are red and

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A certain basket contains 10 apples, 7 of which are red and [#permalink] New post 24 Jan 2011, 00:05
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A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are randomly selected, what is the probability that out of those 3 , 2 will be red and 1 will be green ?

A. 7/40
B. 7/20
C. 49/100
D. 21/40
E. 7/10
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A certain basket contains 10 apples, 7 of which are red and [#permalink] New post 24 Jan 2011, 01:34
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tradinggenius wrote:
A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are randomly selected, what is the probability that out of those 3 , 2 will be red and 1 will be green ?

1) 7/40
2) 7/20
3) 49/100
4) 21/40
5) 7/10


Probability = # of favorable outcomes / total # of outcomes;

\(P=\frac{C^2_7*C^1_3}{C^3_{10}}=\frac{21}{40}\), where \(C^2_7\) is # of ways to choose 2 different green apples out of 7, \(C^1_3\) is # of ways to choose 1 red apple out of 3, and \(C^3_{10}\) is total # of ways to choose 3 different apples out of total 10 apples.

Answer: D.

Or by probability approach: \(P(RRG)=\frac{3!}{2!}*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}=\frac{21}{40}\), we are multiplying by 3!/2! as the case of GGR can occur in 3 ways: GGR, GRG, RGG - # of permutation of 3 letters out of which 2 are identical.

Answer: D.
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Re: GMAT PREP PROBABILITY [#permalink] New post 05 Feb 2011, 06:38
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probability of getting RRG

\(= \frac{7}{10}*\frac{6}{9}*\frac{3}{8} = \frac{7}{40}\)

there are 3 different ways to get 2 red and 1 green {RRG, RGR, GRR}.

so probability = \(\frac{7*3}{40} = \frac{21}{40}\)

Ans: D

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Re: A certain basket contains 10 apples, 7 of which are red and [#permalink] New post 07 Nov 2014, 05:03
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Re: A certain basket contains 10 apples, 7 of which are red and   [#permalink] 07 Nov 2014, 05:03
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